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# Rearrange array such that sum of same indexed elements is atmost K

• Last Updated : 07 Jul, 2021

Given two arrays A[] and B[] consisting of N integers each and an integer K, the task is to rearrange the array B[] such that sum of Ai + Bi is atmost K. If no such arrangement is possible, print -1.

Examples:

Input: A[] = {1, 2, 3, 4, 2}, B[] = {1, 2, 3, 1, 1}, K = 5
Output: 1 3 1 1 2

Input: A[] = {1, 2, 3, 4, 5}, B[] = {2, 3, 4, 5, 6}, K = 6
Output: -1

Approach: The most optimal rearrangement of the array B[] to satisfy the given conditions is to sort the array in descending order.

Proof:

1. Since the sum of every pair of ith indexed elements can be atmost K. Therefore, mn + num ≤ X and mx + num1 ≤ X, where mn and mx are the minimum elements in the array A[] and B[] respectively.
2. Therefore, by induction, it can be proved that mn and mx can be paired.
3. Therefore, sorting the array in descending order is the most optimal rearrangement

Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to rearrange array such``// that sum of similar indexed elements``// does not exceed K``void` `rearrangeArray(``int` `A[], ``int` `B[],``                    ``int` `N, ``int` `K)``{``    ``// Sort the array B[]``    ``// in descending order``    ``sort(B, B + N, greater<``int``>());` `    ``bool` `flag = ``true``;` `    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// If condition fails``        ``if` `(A[i] + B[i] > K) {``            ``flag = ``false``;``            ``break``;``        ``}``    ``}` `    ``if` `(!flag) {` `        ``cout << ``"-1"` `<< endl;``    ``}``    ``else` `{` `        ``// Print the array``        ``for` `(``int` `i = 0; i < N; i++) {``            ``cout << B[i] << ``" "``;``        ``}``    ``}``}` `// Driver Code``int` `main()``{``    ``// Given arrays``    ``int` `A[] = { 1, 2, 3, 4, 2 };``    ``int` `B[] = { 1, 2, 3, 1, 1 };` `    ``int` `N = ``sizeof``(A) / ``sizeof``(A);` `    ``int` `K = 5;` `    ``rearrangeArray(A, B, N, K);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG{` `// Reverse array``static` `int``[] reverse(``int` `a[])``{``    ``int` `i, n = a.length, t;``    ``for``(i = ``0``; i < n / ``2``; i++)``    ``{``        ``t = a[i];``        ``a[i] = a[n - i - ``1``];``        ``a[n - i - ``1``] = t;``    ``}``    ``return` `a;``}``    ` `// Function to rearrange array such``// that sum of similar indexed elements``// does not exceed K``static` `void` `rearrangeArray(``int` `A[], ``int` `B[],``                           ``int` `N, ``int` `K)``{``    ` `    ``// Sort the array B[]``    ``// in descending order``    ``Arrays.sort(B);``    ``B = reverse(B);` `    ``boolean` `flag = ``true``;` `    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ` `        ``// If condition fails``        ``if` `(A[i] + B[i] > K)``        ``{``            ``flag = ``false``;``            ``break``;``        ``}``    ``}` `    ``if` `(!flag)``    ``{``        ``System.out.print(``"-1"` `+ ``"\n"``);``    ``}``    ``else``    ``{``        ` `        ``// Print the array``        ``for``(``int` `i = ``0``; i < N; i++)``        ``{``            ``System.out.print(B[i] + ``" "``);``        ``}``    ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ` `    ``// Given arrays``    ``int` `A[] = { ``1``, ``2``, ``3``, ``4``, ``2` `};``    ``int` `B[] = { ``1``, ``2``, ``3``, ``1``, ``1` `};` `    ``int` `N = A.length;` `    ``int` `K = ``5``;` `    ``rearrangeArray(A, B, N, K);``}``}` `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 program for the above approach` `# Function to rearrange array such``# that sum of similar indexed elements``# does not exceed K``def` `rearrangeArray(A, B, N, K):``    ` `    ``# Sort the array B[]``    ``# in descending order``    ``B.sort(reverse ``=` `True``)` `    ``flag ``=` `True` `    ``for` `i ``in` `range``(N):``        ` `        ``# If condition fails``        ``if` `(A[i] ``+` `B[i] > K):``            ``flag ``=` `False``            ``break` `    ``if` `(flag ``=``=` `False``):``        ``print``(``"-1"``)``    ``else``:``        ` `        ``# Print the array``        ``for` `i ``in` `range``(N):``            ``print``(B[i], end ``=` `" "``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Given arrays``    ``A ``=` `[ ``1``, ``2``, ``3``, ``4``, ``2` `]``    ``B ``=` `[ ``1``, ``2``, ``3``, ``1``, ``1` `]` `    ``N ``=` `len``(A)``    ``K ``=` `5``;` `    ``rearrangeArray(A, B, N, K)` `# This code is contributed by SURENDRA_GANGWAR`

## C#

 `// C# program for the``// above approach``using` `System;``class` `GFG{` `// Reverse array``static` `int``[] reverse(``int` `[]a)``{``  ``int` `i, n = a.Length, t;``  ` `  ``for``(i = 0; i < n / 2; i++)``  ``{``    ``t = a[i];``    ``a[i] = a[n - i - 1];``    ``a[n - i - 1] = t;``  ``}``  ``return` `a;``}``    ` `// Function to rearrange array such``// that sum of similar indexed elements``// does not exceed K``static` `void` `rearrangeArray(``int` `[]A, ``int` `[]B,``                           ``int` `N, ``int` `K)``{   ``  ``// Sort the array []B``  ``// in descending order``  ``Array.Sort(B);``  ``B = reverse(B);` `  ``bool` `flag = ``true``;` `  ``for``(``int` `i = 0; i < N; i++)``  ``{``    ``// If condition fails``    ``if` `(A[i] + B[i] > K)``    ``{``      ``flag = ``false``;``      ``break``;``    ``}``  ``}` `  ``if` `(!flag)``  ``{``    ``Console.Write(``"-1"` `+ ``"\n"``);``  ``}``  ``else``  ``{``    ``// Print the array``    ``for``(``int` `i = 0; i < N; i++)``    ``{``      ``Console.Write(B[i] + ``" "``);``    ``}``  ``}``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{   ``  ``// Given arrays``  ``int` `[]A = {1, 2, 3, 4, 2};``  ``int` `[]B = {1, 2, 3, 1, 1};` `  ``int` `N = A.Length;``  ``int` `K = 5;``  ``rearrangeArray(A, B, N, K);``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``
Output:
`3 2 1 1 1`

Time Complexity: O(N)
Auxiliary Space Complexity: O(1)

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